Let's See if You Have Geometry Reasoning

Let [ABC] denote the area of ABC

The diagonal PR divides the entire rectangle in half: [PSR] = [PQR]

So, [PEI] + [ESHI] + [IHR] = [PFI] + [FQGI] + [GIR]

The diagonal also divides two smaller shaded regions in half: [PEI] = [PFI] And [GIR] = [IHR]

It follows that [ESHI] = [FQGI] Or [A] = [B]

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See Mark's image for the notation to use.

• Note that $\mbox{PE}=\mbox{IF}$ and $\mbox{GI}=\mbox{HR}$ .
• Note that the triangles $\mbox{PEI}$ and $\mbox{IHR}$ are similar.
• Therefore, $\dfrac{\mbox{PE}}{\mbox{EI}}=\dfrac{\mbox{IH}}{\mbox{HR}}$
• By above, $\dfrac{\mbox{IF}}{\mbox{EI}}=\dfrac{\mbox{IH}}{\mbox{GI}}$
• By cross multiplication , $\mbox{IF}\times \mbox{GI}=\mbox{IH}\times \mbox{EI}$ , which is the result wanted,
• since the area of $\mbox A$ is $\mbox{IH}\times \mbox{EI}$ ,
• and the area of $\mbox B$ is $\mbox{IF}\times \mbox{GI}$ .

Any diagonal of a rectangle divides it in half. Therefore, the area of PRS equals the area of PRQ; and the area of PIE equals the area of PIF and the area of IGR equals the area of IHR. Since PQR is equal to PRS and if you remove the above noted equal areas from each, then the two areas left, A and B, must be equal.

Please refer to the figure shown by Mark Mottian: PEI+A+RHI = PFI+B+RGI Equation (1) Since: PEI = PFI Equation (2) RHI = RGI Equation (3) Substitute Equation (2) & (3) in Equation (1`) PEI + A + RHI = PEI + B + RHI Cancel: PEI, RHI Therefore: A = B

If we try to visualise triangle psr and pqr can be proved congurent by SAS and the triangles forming within these triangle (unshaded area )Are congurent now if equal are if taken off from congurent triangle the left area if also equal . We can prove the congruence easily and it is visible very much i hope i dont need to explain that

In a short, rough, unofficial, initial analysis, I have concluded that: EI > FI > IG > IH (use sir Mark Mottian's image of solution as reference since I have only judged it mentally, so, there is no drawing / image to upload.) Then, the two rectangles are being separated by a diagonal inside a rectangle. Area of Rectangle EISH = (EI)(IH) and Area of Rectangle FQIG = (FI)(IG). Rectangle EISH uses the longest and the shortest division of the proportion (Extremes), and Rectangle FQIG uses the 2nd and 3rd longest division (Means), so I have concluded that they must be equal. (Though I still tried to properly solve it using ratio and proportion like what sir Mark did to confirm if I am indeed correct, and, I am.) This is only possible because the two rectangles were separated by the diagonal inside of the biggest rectangle. I apologize for my bad and confusing grammar, though.

No trig PSR and trig POR will be equal. the left over traingles wil be same area , hence A =B

See Mark's image for the notation to use. Consider rectangle IGRH, and let that have area C.

By similar triangles, $\frac{IH}{PS} = \frac{ HR}{SR}$ .

Hence, $\frac{C}{A} = \frac{ HR}{SR} = \frac{ IH}{FH} = \frac{C}{B}$ .

Thus, $A = B$ .